Kết quả nè bạn ơi;

A; 3/4 + 2/5=23/20

B;7/4 - 1/2=5/4

C;4/9 x 6/8=24/72

D;14/5 : 7/8=112/35

OK bạn đợi mình tý

bạn giúp mik bằng cách: ghi rõ phép tính nha 

cảm ơn 

1+2+3+4+5+6+7+8+9+10=55       Ko chắc:=))

bằng 55

BN Nguyễn Ngọc Diệp sai r = 55 mà

a)\(\frac{3}{4}+\frac{2}{5}=\frac{15}{20}+\frac{8}{20}=\frac{23}{20}\)

b) \(\frac{7}{4}-\frac{1}{2}=\frac{14}{8}-\frac{4}{8}=\frac{10}{8}=\frac{5}{4}\)

c) \(\frac{4}{9}.\frac{6}{8}=\frac{24}{72}=\frac{1}{3}\)

d) \(\frac{14}{5}:\frac{7}{8}=\frac{14}{5}.\frac{8}{7}=\frac{72}{35}\)

a)\(x^3y+3x^2y\)

b)\(-24x^2+4xy\)

[9x3(x2-1)-6x2(x2-1)2+12x(x2-1)]:3x(x2-1)

Gọi x2-1 là y tá đc

(9x3y-6x2y2+12xy):3xy

=3x2-2xy+4

=3x2-2x(x2-1)+4

a) \(-\frac{1}{15}-\frac{1}{16}=-\frac{16}{240}-\frac{15}{240}=-\frac{31}{240}\)

b) \(\frac{7}{24}-\left(-\frac{5}{36}\right)=\frac{7}{24}+\frac{5}{36}=\frac{21}{72}+\frac{10}{72}=\frac{31}{72}\)

c) \(-\frac{7}{9}-\left(-\frac{7}{12}\right)=-\frac{7}{9}+\frac{7}{12}=-\frac{28}{36}+\frac{21}{36}=-\frac{7}{36}\)

\(a)\frac{-1}{15}-\frac{1}{16}\)

\(\frac{-1}{15}-\frac{1}{16}.MSC:240\)

\(=\frac{-1.16}{15.16}-\frac{1.15}{16.15}=\frac{-16}{240}-\frac{15}{240}=\frac{-16-15}{240}=\frac{-31}{240}\)

\(b)\frac{7}{24}-\frac{-5}{36}\)

\(\frac{7}{24}-\frac{-5}{36}.MSC:72\)

\(=\frac{7.3}{24.3}-\frac{-5.2}{36.2}=\frac{21}{72}-\frac{-10}{72}=\frac{21-\left(-10\right)}{72}=\frac{31}{72}\)

\(c)\frac{-7}{9}-\frac{-7}{12}\)

\(\frac{-7}{9}-\frac{-7}{12}.MSC:36\)

\(=\frac{-7.4}{9.4}-\frac{-7.3}{12.3}=\frac{-28}{36}-\frac{-21}{36}=\frac{-28-\left(-21\right)}{36}=\frac{-7}{36}\)

   Chúc bạn một ngày vui vẻ ~! ❤‿❤

    Nếu có gì sai thì báo lỗi cho mình biết nhé.

-(-x)/5-2/10-1/-5-7/50

=>-(-x)/5-2/10=-17/50

=>-(-x)/5=-7/50

=>-(-x)=-7/10

=>x=-7/10

2.x+10/3=22/3

2.x=22/3-10/3

2.x=4

x=4:2

x=2

vậy x=2

Trả lời:

\(2x+\frac{31}{3}=7\frac{1}{3}\)

\(\Leftrightarrow2x+\frac{31}{3}=\frac{22}{3}\)

\(\Leftrightarrow2x=\frac{-9}{3}\)

\(\Leftrightarrow2x=-3\)

\(\Leftrightarrow x=\frac{-3}{2}\)

Vậy \(x=\frac{-3}{2}\)

Trả lời:

\(2x+3\frac{1}{3}=7\frac{1}{3}\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

phá ngoặc ra mà tính đi bn ơi

a. \(\left(x^2-2x+1\right)-3x\left(x-1\right)=0\)

\(\Leftrightarrow x^2-2x+1-3x^2+3x=0\)

\(\Leftrightarrow-2x^2+x+1=0\)

\(\Leftrightarrow-2x^2+2x-x+1=0\)

\(\Leftrightarrow-2x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow-\left(2x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)

Vậy \(x\in\left\{-\frac{1}{2};1\right\}\)

b. \(4\left(7x-3\right)-\left(7x^2-3x\right)=0\)

\(\Leftrightarrow4\left(7x-3\right)-x\left(7x-3\right)=0\)

\(\Leftrightarrow\left(4-x\right)\left(7x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4-x=0\\7x-3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{3}{7}\end{cases}}\)

Vậy \(x\in\left\{4;\frac{3}{7}\right\}\)

c.\(\left(5-x\right)\left(2+3x\right)=4-9x^2\)

\(\Leftrightarrow\left(5-x\right)\left(2+3x\right)=\left(2-3x\right)\left(2+3x\right)\)

\(\Leftrightarrow\left(2+3x\right)\left(5-x-2+3x\right)=0\)

\(\Leftrightarrow\left(2+3x\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2+3x=0\\2x+3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=-\frac{3}{2}\end{cases}}\)

Vậy \(x\in\left\{-\frac{2}{3};-\frac{3}{2}\right\}\)

d. \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow7-2x-4=-x-4\)

\(\Leftrightarrow7-4+4=-x+2x\)

\(\Leftrightarrow7=x\)

Vậy x = 7

e. \(\left(x-1\right)-\left(2x-1\right)=9\)

\(\Leftrightarrow x-1-2x+1=9\)

\(\Leftrightarrow-x=9\)

\(\Leftrightarrow x=-9\)

g. \(x^3+x^2+x+1=0\)

\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x+1=0\end{cases}}\)Mà : \(x^2+1\ge1>0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy x = -1

\(2x^2-4x-5=2x^2-4x+2-7=2\left(x-1\right)^2-7\ge0-7=-7\Leftrightarrow x=1\)

\(-2x^2-6x+15=-2x^2-6x-4,5+19,5=-2\left(x+\frac{3}{2}\right)^2+19,5\le0+19,5=19,5\Leftrightarrow x=\frac{-3}{2}\)

Bài 1 : Tìm giá trị lớn nhất, nhỏ nhất

a, \(2x^2-4x-5=2\left(x^2-2x+1\right)-7=2\left(x-1\right)^2-7\)

Vì \(2\left(x-1\right)^2\ge0\Rightarrow2x^2-4x-5\ge-7\)

\(''=''\Leftrightarrow x=1\)

b, \(-2x^2-6x+15=-2\left(x^2+2x.\frac{3}{2}+\frac{9}{4}\right)+\frac{39}{2}=-2\left(x+\frac{3}{2}\right)^2+\frac{39}{2}\)

Vì \(-2\left(x+\frac{3}{2}\right)^2\le0\Rightarrow-2x^2-6x+15\le\frac{39}{2}\)

\(''=''\Leftrightarrow x=-\frac{3}{2}\)

Bài 2 : Tìm x

a, \(2x^3-3x^2+2=0\) (tạm thời chưa ra)

b, \(x^4-2x^2+1=0\)

\(\Leftrightarrow\left(x^2-1\right)^2=0\Rightarrow x^2-1=0\Rightarrow x=\pm1\)

b) x\(^4\) - 2x\(^2\) + 1 = 0

⇔ ( x\(^2\) - 1 )\(^2\) = 0

⇔ x\(^2\) - 1 = 0

⇔ x\(^{ }\) = \(\pm\) 1